Many-Electron AtomsWe have to examine the balance of attractions and repulsions in the

atom to explain why subshells of a given shell have different energies. As well as being attracted by the nucleus, each electron in a many-electron atom is repelled by the other electrons present. As a result, it is less tightly bound to the nucleus than it would be if those other electrons were absent. We say that each electron is shieldedshielded from the full attraction of the nucleus by the other electrons in the

atom. The shielding effectively reduces the pull of the nucleus on an electron. The effective nuclear charge, Zeffective nuclear charge, Zeffeff,, experienced by the electron is always less than the actual nuclear charge, Z, because the electron-electron repulsions work against the pull of the nucleus. Note that the other electrons do not “block” the influence of the nucleus; they simply provide additional repulsive Coulombic interactions that partly counteract the pull of the nucleus. Finally, an s-electron of any shell can be found very close to the nucleus, so we say that it can penetratepenetrate through the inner shells. A p-electron penetrates much less. Because a p-electron penetrates less than an s-electron through the inner shells of the atom, it is more effectively shielded from the nucleus and hence experiences a smaller effective nuclear charge than an s-electron does. That is, an s-electron is bound more tightly than a p-electron and has a slightly lower (more negative) energy. In a many-electron atom, because of the effects of penetration and shielding, the order of energies of orbitals in a given shell is typically ss < < pp < < dd < < ff..

A few principles to consider:

Aufbau Principle: The procedure for arriving at the ground-state electron configurations of atoms and molecules in order of increasing atomic number. To proceed from one atom to the next, we add a proton and some neutrons to the nucleus and then describe the orbital into which the added electron goes.

Hund’s Rule: Whenever orbitals of equal energy (degenerate) are available, electrons occupy these orbitals singly before pairing begins.

Core vs. Valence electrons: inner vs. outermost electrons (latter contained within outermost shell)

Electron ConfigurationElectron configuration is a shorthand

notation for describing the arrangement of the electrons about

the nucleus.General Format using the quantum numbers:

n n l l e-e-

RULES:

1. Fill the lowest energy levels first.

1s 2s 2p 3s 3p 4s 3d 4p

2. No more than two electrons per orbital.

n = principle quantum number

l = angular momentum quantum number

e- = number of electrons

Lowest

Electron ConfigurationExamples:Examples:

HH :: 1s1 He:He: 1s2 Li :Li : 1s2 2s1

Co: 1s1s22 2s 2s22 2p 2p66 3s 3s22 3p 3p6 6 4s4s22 3d 3d77

Br: 1s1s22 2s 2s22 2p 2p66 3s 3s22 3p 3p6 6 4s4s22 3d 3d10 10 4p4p55

The condensed electron configuration distinguishes the core electrons from the valence electrons. CORE CORE electrons are tightly held to the nucleus and resemble a noble gas configuration. VALENCE VALENCE electrons are the outer most electrons and are involved in chemical

reactions. Examples of the condensed configuration:

Li:[He] 2s1 Co:[Ar] 4s2 3d3d7 7

Br:[Ar] 4s2 3d3d10 10 4p4p55

Electron ConfigurationThe full & condensed electron configuration for some elements:

CC 1s 1s22 2s 2s22 2p 2p22 oror [He] 2s[He] 2s22 2p 2p22

OO 1s 1s22 2s 2s22 2p 2p44 or or [He] 2s[He] 2s22 2p 2p22

NeNe 1s 1s22 2s 2s22 2p 2p66 or or [Ne][Ne]

NaNa 1s 1s22 2s 2s22 2p 2p66 3s 3s11 oror [Ne] 3s[Ne] 3s11

Si Si 1s1s22 2s 2s22 2p 2p66 3s 3s22 3p 3p22 oror [Ne] 3s[Ne] 3s22 3p 3p22

ClCl 1s 1s22 2s 2s22 2p 2p66 3s 3s22 3p 3p55 oror [Ne] 3s[Ne] 3s22 3p 3p55

ArAr 1s 1s22 2s 2s22 2p 2p66 3s 3s22 3p 3p66 oror [Ar][Ar]

K K 1s1s22 2s 2s22 2p 2p66 3s 3s22 3p 3p6 6 4s4s11 oror [Ar]4s[Ar]4s11

7

s1

s2

d1 d2 d3 d4 d5 d6 d7 d8 d9 d10

s2p1 p2 p3 p4 p5

p6

f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f14d1

1234567

8

Electron Configuration fromthe Periodic Table

P = [Ne]3s23p3

P has 5 valence electrons

3p3

P

Ne1234567

1A

2A 3A 4A 5A 6A 7A

8A

3s2

9

Sublevel Splitting in Multielectron Atoms

• the sublevels in each principal energy level of Hydrogen all have the same energy – we call orbitals with the same energy degenerate– or other single electron systems

• for multielectron atoms, the energies of the sublevels are split– caused by electron-electron repulsion

• the lower the value of the l quantum number, the less energy the sublevel has– s (l = 0) < p (l = 1) < d (l = 2) < f (l = 3)

10

Penetrating and Shielding• the radial distribution function shows that

the 2s orbital penetrates more deeply into the 1s orbital than does the 2p

• the weaker penetration of the 2p sublevel means that electrons in the 2p sublevel experience more repulsive force, they are more shielded from the attractive force of the nucleus

• the deeper penetration of the 2s electrons means electrons in the 2s sublevel experience a greater attractive force to the nucleus and are not shielded as effectively

• the result is that the electrons in the 2s sublevel are lower in energy than the electrons in the 2p

11

Penetration & Shielding

Ene

rgy

1s

7s

2s

2p

3s

3p3d

6s6p

6d

4s

4p4d

4f

5s

5p

5d5f

Notice the following:1. because of penetration, sublevels within

an energy level are not degenerate2. penetration of the 4th and higher energy

levels is so strong that their s sublevel is lower in energy than the d sublevel of the previous energy level

3. the energy difference between levels becomes smaller for higher energy levels

QUANTUM MECHANICS & ORBITAL DIAGRAMSOrbital Energy Levels:

_ _ _ 6p _ _ _ _ _

5d _ _ _ _ _ _ ___ 4f

E 6s _ _ _n 5p _ _ _ _ _e 4dr __g 5s _ _ _y 4p _ _ _ _ _

3d__4s _ _ _

3p__3s _ _ _

2p__2s__

1s

Example of Ionization Energies:

Al(g) Al+(g) + e- I1 = 580 kJ/mol

Al+(g) Al2+

(g) + e- I2 = 1815 kJ/molAl2+

(g) Al3+(g) + e- I3 = 2740 kJ/mol

Al3+(g) Al4+

(g) + e- I4 = 11,600 kJ/mol

Orbital DiagramsOrbital DiagramsOrbital diagrams are written in order of increasing energy

levels starting with the lowest energy level the 1s orbital.

___ ___ ___ 4p

___ ___ ___ ___ ___ 3d

___ 4s

___ ___ ___ 3p

___ 3s

___ ___ ___ 2p

___ 2s

___ 1s

RULES:

(1) fill the lowest energy level first

(2) fill each orbital in a subshell with one electron first before you double up.

(3) Completely fill each subshell before proceeding to the next energy level.

Remember the Remember the order!!order!!

Orbital DiagramsOrbital DiagramsFill in the orbital diagrams for:

C C OO

___ ___ ___ 4p ___ ___ ___ 4p

__ __ __ __ __ 3d __ __ __ __ __ 3d

___ 4s ___ 4s

___ ___ ___ 3p ___ ___ ___ 3p

___ 3s ___ 3s

___ ___ ___ 2p ___ ___ ___ 2p

___ 2s ___ 2s

___ 1s ___ 1s

16

Transition Elements• for the d block metals, the principal energy level is one less than

valence shell– one less than the Period number– sometimes s electron “promoted” to d sublevel

4s 3d

ZnZ = 30, Period 4, Group 2B[Ar]4s23d10

• for the f block metals, the principal energy level is two less than valence shell two less than the Period number they really belong tosometimes d electron in configurationEuZ = 63, Period 6[Xe]6s24f 7 6s 4f

17

Electron Configuration fromthe Periodic Table

As = [Ar]4s23d104p3

As has 5 valence electrons

As

1234567

1A

2A 3A 4A 5A 6A 7A

8A

4s2

Ar3d10

4p3

18

Practice – Use the Periodic Table to write the short electron configuration and orbital diagram for each

of the following

• Na (at. no. 11)

• Te (at. no. 52)

• Tc (at. no. 43)

19

Practice – Use the Periodic Table to write the short electron configuration and orbital diagram for each

of the following

• Na (at. no. 11) [Ne]3s1

• Te (at. no. 52) [Kr]5s24d105p4

• Tc (at. no. 43) [Kr]5s24d5

3s

5s 5p4d

5s 4d

1. Determine the ground-state electron configuration for each of the following elements:

A. sulfur B. polonium

2. Predict the number of valence electrons present in each of the following atoms (include the outermost d-electrons when necessary):

A. B B. Ba C. Bi

3. Determine the ground-state electron configuration for each of the following ions:

A. Al+3 B. Tc+4

4. Predict the number of valence electrons present for each of the following ions:

A. In+ B. Tc+2

5. Give the ground-state electron configuration and number of unpaired electrons expected for each of the following ions:

A. Ga3+ B. Cu+2

6. For each of the following ground-state ions, predict the type of orbital that the electrons of highest energy will occupy:

A. Fe+2 B. Bi+3

Lecture Questions

Workshop on electron configuration1. Determine the ground-state electron configuration for each of the following elements (see last page of this section for sample energy levels):

A. chlorine B. cesiumC. vanadium D. rhenium

2. Predict the number of valence electrons present in each of the following atoms (include the outermost d-electrons):

A. Sn B. La C. Mn D. Zn3. Determine the ground-state electron configuration for each of the following ions:

A. Co+3B. Mo+2 C. Ra+2

D. I- E. Ir+ F. Ru+4

4. Predict the number of valence electrons present for each of the following ions:A. Tl+ B. Po+2 C. Ta+2 D. Re+

5. Give the ground-state electron configuration and number of unpaired electrons expected for each of the following ions:

A. Ga+ B. Cu+1C. Pb+2 D. Se-2

6. For each of the following ground-state ions, predict the type of orbital that the electrons of highest energy will occupy:

A. Fe+3 B. B+3 C. As+3 D. Os+

PERIODICITYDiamagnetic vs. Paramagnetic species:

Diamagnetic has all its electrons paired and is slightly repelled by a magnetic field

Paramagnetic has one or more unpaired electrons and is attracted into a magnetic field.

Which group(s) on the periodic table will have elements that are always diamagnetic?

Periodic Trends1. Atomic Radius

2. Ionization Energy – energy needed to remove an electron from gaseous atom

Decreases acrossincreases

Increases down

Increases acrossDecreases down

3. Electron Affinity – energy released when an electron is added to gaseous atom

4. Electronegativity – the electron pulling power of an atom when it is part of a molecule (denoted with the Greek letter )

5. Metallic Character

Workshop on periodic trends1. Arrange the following in terms of DECREASING atomic radius & then first ionization energy & then electronegativity:

Be, B, C, N, O, F, Ne

2. Why is the first ionization energy of aluminum slightly lower than the first ionization energy for magnesium?

3. Why is the second ionization energy for sodium so much greater than its first ionization energy?

4. Arrange the following in terms of DECREASING atomic (or ionic) radii: O+, O, O-

5. Give a reason why the electronegativity for F is so much greater than the electronegativity for Fr.